Proton Suspension Between Horizontal Parallel Plates: Charge & Voltage

[lizaliiu]

1. A proton is suspended in a vacuum between two oppositely charged horizontal parallel plates. The separation between the plates is 2.60mm



2. I assume it uses E=vd/d, but I don't really know



3. what is the sign of the charge on the lower plate?
what is the voltage across the plate?

 

[mfb]

2. I assume it uses E=vd/d, but I don't really know

What does that mean?

Some hints how to start:
Can you identify the forces acting on the proton?
If the proton stays at rest between the plates, can you identify the direction of those forces?

 

[berkeman]

1. A proton is suspended in a vacuum between two oppositely charged horizontal parallel plates. The separation between the plates is 2.60mm



2. I assume it uses E=vd/d, but I don't really know



3. what is the sign of the charge on the lower plate?
what is the voltage across the plate?

Welcome to the PF.

You are required to show your work toward a solution before we can offer much tutorial help.

 

[lizaliiu]

since gravity pulls the proton downward so the lower plate should be the positive charge to repel the proton.
F=mg=qE
E=mg/q=v/d, m=1.673x10^-27kg, g=9.8m/s2, q=1.602x10^-19 and d=2.60x10^-3m
so we just need to solve q to get the voltage across the plates?

 

[mfb]

You know q, I think you mean V? Right.

 

[lizaliiu]

yes, I solved V= 2.66x10^-10...not positive sure it's correct tho

 

[mfb]

Gravity is weak, and you can check your calculations in WolframAlpha, for example. If the units match, it should be right.